The function f, defined by ` f(x)=(x^(2))/(2) +In x - 2 cos x ` increases for ` x in `

Options

A. `R^(-)`
B. `R^(+)`
C. `R-{0}`
D. `[1,oo)`

Answer

To determine the intervals where the function \( f(x) = \frac{x^2}{2} + \ln x - 2 \cos x \) is increasing, we need to find the derivative \( f'(x) \) and analyze where it is greater than or equal to zero.

### Step 1: Differentiate the function
We start by differentiating \( f(x) \):

\[ f'(x) = \frac{d}{dx}\left(\frac{x^2}{2}\right) + \frac{d}{dx}(\ln x) - \frac{d}{dx}(2 \cos x) \]

Calculating each term:

1. The derivative of \( \frac{x^2}{2} \) is \( x \).
2. The derivative of \( \ln x \) is \( \frac{1}{x} \).
3. The derivative of \( -2 \cos x \) is \( 2 \sin x \).

Thus, we have:

\[ f'(x) = x + \frac{1}{x} + 2 \sin x \]

### Step 2: Set the derivative greater than or equal to zero
To find where the function is increasing, we need to solve the inequality:

\[ f'(x) \geq 0 \]

This translates to:

\[ x + \frac{1}{x} + 2 \sin x \geq 0 \]

### Step 3: Analyze the inequality
We can rewrite the inequality as:

\[ x + 2 \sin x + \frac{1}{x} \geq 0 \]

### Step 4: Consider the domain of \( f(x) \)
Since \( \ln x \) is defined for \( x > 0 \), we restrict our analysis to \( x > 0 \).

### Step 5: Check the components of the inequality
1. The term \( x + \frac{1}{x} \) is always positive for \( x > 0 \) since both \( x \) and \( \frac{1}{x} \) are positive.
2. The term \( 2 \sin x \) oscillates between -2 and 2.

Thus, we need to check when:

\[ x + \frac{1}{x} + 2 \sin x \geq 0 \]

### Step 6: Analyze the critical points
- The minimum value of \( x + \frac{1}{x} \) occurs at \( x = 1 \), where it equals 2.
- Since \( 2 \sin x \) can take values from -2 to 2, we find that:

\[ x + \frac{1}{x} \geq 0 \text{ for all } x > 0 \]

### Step 7: Conclusion
Since \( x + \frac{1}{x} \) is always positive and can counteract the negative values of \( 2 \sin x \) for all \( x > 0 \), we conclude that:

The function \( f(x) \) is increasing for all \( x > 0 \).

### Final Answer:
The function \( f(x) \) increases for \( x \in (0, \infty) \).

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